Definitions of hyperbolic functions in terms of exponentials, their graphs and identities, inverse hyperbolic functions in logarithmic form, and differentiation and integration involving them.

What this dot point is asking

AQA wants you to define the hyperbolic functions from exponentials, sketch their graphs, use the hyperbolic identities, write the inverse hyperbolic functions in logarithmic form, and differentiate and integrate expressions involving them, including the standard integrals that produce inverse hyperbolic functions.

Definitions and graphs

The graph of $\sinh x$ rises through the origin with no turning points, $\cosh x$ is a symmetric U-shape with a minimum of $1$ at the origin (the shape of a hanging chain, the catenary), and $\tanh x$ is the bounded S-curve squeezed between the asymptotes $y = -1$ and $y = 1$.

Identities and Osborn's rule

The key identity is $\cosh^2 x - \sinh^2 x = 1$. Osborn's rule says you can adapt a trig identity to a hyperbolic one by changing the sign of any product (or implied product) of two sines. So $\cos^2 + \sin^2 = 1$ becomes $\cosh^2 - \sinh^2 = 1$.

Inverse hyperbolic functions

Because the hyperbolics are built from exponentials, their inverses are logarithms.

Solving hyperbolic equations

Many exam problems reduce to a polynomial in $\sinh x$ or $\cosh x$ via the identity $\cosh^2 x - \sinh^2 x = 1$. Replace one squared function in terms of the other, form a quadratic, solve it, then convert back using the logarithmic inverse forms. Watch the domains: $\sinh x$ takes every real value, so any real solution for $\sinh x$ is valid, but $\cosh x \geq 1$, so a value of $\cosh x$ below $1$ must be rejected.

Differentiation and integration

The derivatives are $\frac{d}{dx}\sinh x = \cosh x$, $\frac{d}{dx}\cosh x = \sinh x$ and $\frac{d}{dx}\tanh x = \operatorname{sech}^2 x$, all following directly from the exponential definitions. Reversing them, together with the inverse forms, gives the two standard integrals examined most: $\int \frac{1}{\sqrt{x^2 + 1}}\,dx = \operatorname{arsinh} x + c$ and $\int \frac{1}{\sqrt{x^2 - 1}}\,dx = \operatorname{arcosh} x + c$. With a coefficient on $x^2$, factor it out of the surd first to reach the standard form $\sqrt{x^2 + a^2}$ or $\sqrt{x^2 - a^2}$, which produces $\operatorname{arsinh}\frac{x}{a}$ or $\operatorname{arcosh}\frac{x}{a}$.

To verify the core identity from first principles, expand the exponential definitions: $\cosh^2 x - \sinh^2 x = \frac{1}{4}(e^x + e^{-x})^2 - \frac{1}{4}(e^x - e^{-x})^2$. The squared terms $e^{2x}$ and $e^{-2x}$ cancel between the two brackets, and the cross terms leave $\frac{1}{4}(2 + 2) = 1$. This identity, the logarithmic inverses, and the parallel-to-trig derivatives are the three tools a hyperbolic question almost always combines, so keep the sign differences from the circular case firmly in mind.